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2 years ago

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A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.3

80 m/s^2. Suppose it moves with constant acceleration for 3.80 s.
Required:
a. Find the position of the particle after this time.
b. Find its velocity at the end of this time interval.

1 answer:

shusha [124]2 years ago

6 0

Answer:

a

$x_2 = -2.3356$

b

$v = -1.384 \ m/s$

Explanation:

From the question we are told that

The initial position of the particle is $x_1 = 0.180 \ m$

The initial velocity of the particle is $u = 0.060 \ m/s$

The acceleration is $a = -0.380 \ m/s^2$

The time duration is $t = 3.80 \ s$

Generally from kinematic equation

$v = u + at$

=> $v = 0.060 + (-0.380 * 3.80)$

=> $v = -1.384 \ m/s$

Generally from kinematic equation

$v^2 = u^2 + 2as$

Here s is the distance covered by the particle, so

$(-1.384)^2 = (0.060)^2 + 2(-0.380)* s$

=> $s = -2.5156 \ m$

Generally the final position of the particle is

$x_2 = x_1 + s$

=> $x_2 = 0.180 + (-2.5156)$

=> $x_2 = -2.3356$

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